Solving equations involving exponents and logarithms

Lets review some terms.

When we write log5 125

5 is called the base125 is called the argument

Logarithmic form of 52 = 25 is

log525 = 2

For all the laws a, M and N > 0a 1r is any real

Remember ln and logln is a short cut for logelog means log10

Log laws

If your variable is in an exponent or in the argument of a logarithmFind the pattern your equation resembles

If your variable is in an exponent or in the argument of a logarithmFind the pattern Fit your equation to match the patternSwitch to the equivalent formSolve the resultCheck (be sure you remember the domain of a log)

log(2x) = 3

It fits log(2x) = 3

Switch log(2x) = 3103=2xDid you remember that log(2x) means log10(2x)?

Divide by 2 log(2x) = 3103=2x500 = x

ln(x+3) = ln(-7x)

ln(x+3) = ln(-7x) It fits

ln(x+3) = ln(-7x)Switch

ln(x+3) = ln(-7x)x + 3 = -7x

Switch

ln(x+3) = ln(-7x)x + 3 = -7xx = -

Solve the result(and check)

ln(x) + ln(3) = ln(12)

ln(x) + ln(3) = ln(12)x + 3 = 12

ln(x) + ln(3) = ln(12)x + 3 = 12 Oh NO!!! Thats wrong!

You need to use log lawsln(x) + ln(3) = ln(12) ln(3x) = ln (12)

Switchln(x) + ln(3) = ln(12)ln(3x) = ln (12)3x = 12

ln(x) + ln(3) = ln(12)ln(3x) = ln (12)3x = 12x = 4Solve the result

log3(x+2) + 4 = 9

It will fit log3(x+2) + 4 = 9

Subtract 4 to make it fit log3(x+2) + 4 = 9log3(x+2) = 5

Switch log3(x+2) + 4 = 9log3(x+2) = 5

Switch log3(x+2) + 4 = 9log3(x+2) = 535 = x + 2

Solve the resultlog3(x+2) + 4 = 9log3(x+2) = 535 = x + 2 x = 241

5(10x) = 19.45

Divide by 5 to fit5(10x) = 19.4510x = 3.91

Switch 5(10x) = 19.4510x = 3.91

Switch 5(10x) = 19.4510x = 3.91log(3.91) = x

Exact log(3.91) Approx 0.5925(10x) = 19.4510x = 3.91log(3.91) = x 0.592

2 log3(x) = 8

It will fit 2 log3(x) = 8

Divide by 2 to fit 2 log3(x) = 8log3(x) = 4

Switch 2 log3(x) = 8log3(x) = 4

Switch 2 log3(x) = 8log3(x) = 434=x

Then Simplify2 log3(x) = 8log3(x) = 434=x x = 81

log2(x-1) + log2(x-1) = 3

Need to use a log lawlog2(x-1) + log2(x-1) = 3

log2(x-1) + log2(x+1) = 3log2{(x-1)(x+1)} = 3

Switchlog2(x-1) + log2(x+1) = 3log2{(x-1)(x+1)} = 3

Switchlog2(x-1) + log2(x+1) = 3log2{(x-1)(x+1)} = 323=(x-1)(x+1)

and finishlog2(x-1) + log2(x+1) = 3log2{(x-1)(x+1)} = 323=(x-1)(x+1) = x2 -1x = +3 or -3

But -3 does not check!log2(x-1) + log2(x+1) = 3log2{(x-1)(x+1)} = 323=(x-1)(x+1) = x2 -1x = +3 or -3

Exclude -3 (it would cause you to have a negative argument)log2(x-1) + log2(x+1) = 3log2{(x-1)(x+1)} = 323=(x-1)(x+1) = x2 -1x = +3 or -3

Theres more than one way to do this

Can you find why each step is valid?

rules of exponents multiply both sides by 2

- 3 to get exact answerApproximate answer

Heres another way to solve the same equation.

exclude 2nd resultSquare both sidesSimplify

52x - 5x 12 = 0

Factor it. Think of y2 - y-12=052x - 5x 12 = 0(5x 4)(5x + 3) = 0

Set each factor = 052x - 5x 12 = 0(5x 4)(5x + 3) = 05x 4 = 0 or 5x + 3 = 0

Solve first factors equationSolve 5x 4 = 0 5x = 4log54 = x

Solve other factors equationSolve 5x + 3 = 0 5x = -3log5(-3) = x

Oops, we cannot have a negative argumentSolve 5x + 3 = 0 5x = -3log5(-3) = x

Exclude this solution.Only the other factors solution worksSolve 5x + 3 = 0 5x = -3log5(-3) = x

4x+2 = 5x

If M = N then ln M = ln N

4x+2 = 5x

If M = N then ln M = ln N4x+2 = 5x ln(4x+2) = ln(5x )

4x+2 = 5x ln(4x+2) = ln(5x)

4x+2 = 5xln(4x+2) = ln(5x)(x+2)ln(4) =(x) ln(5 )

Distribute 4x+2 = 5xln(4x+2) = ln(5x)(x+2)ln(4) =(x) ln(5 )x ln (4) + 2 ln(4) = x ln (5)

Get x terms on one side 4x+2 = 5x ln(4x+2) = ln(5x)(x+2)ln(4) =(x) ln(5 )x ln (4) + 2 ln(4) = x ln (5)x ln (4) - x ln (5)= 2 ln(4)

Factor out x 4x+2 = 5xln(4x+2) = ln(5x)(x+2)ln(4) =(x) ln(5 )x ln (4) + 2 ln(4) = x ln (5)x ln (4) - x ln (5)= 2 ln(4) x [ln (4) - ln (5)]= 2 ln(4)

Divide by numerical coefficient 4x+2 = 5xln(4x+2) = ln(5x)(x+2)ln(4) =(x) ln(5 )x ln (4) + 2 ln(4) = x ln (5)x ln (4) - x ln (5)= 2 ln(4) x [ln (4) - ln (5)]= 2 ln(4)